Why are equilibrium constants dimensionless?

Apr 14, 2015

An equilibrium constant for liquid-solid solutions is defined as:

${K}_{e q} = \frac{{\left[Y\right]}^{{\nu}_{Y}} {\left[Z\right]}^{{\nu}_{Z}}}{{\left[A\right]}^{{\nu}_{A}} {\left[B\right]}^{{\nu}_{B}}}$
Where A and B are reactants, Y and Z are products, and the ${\nu}_{i}$ are the stoichiometric coefficients.

If there are the same number of moles of products as there are for reactants, the stoichiometric coefficients add up to the same number. In other words:

${\nu}_{Y} + {\nu}_{Z} = {\nu}_{A} + {\nu}_{B}$

As a result, your units are:

$\left[\frac{{M}^{{\nu}_{Y}} \cdot {M}^{{\nu}_{Z}}}{{M}^{{\nu}_{A}} \cdot {M}^{{\nu}_{B}}}\right] = \left[{M}^{{\nu}_{Y} + {\nu}_{Z}} / {M}^{{\nu}_{A} + {\nu}_{B}}\right] = \left[{M}^{{\nu}_{A} + {\nu}_{B}} / {M}^{{\nu}_{A} + {\nu}_{B}}\right] =$ [no units]

So this is only true if the moles of products equal the moles of reactants.

Apr 14, 2015

Double check this Truong-Son. The number of moles of reactants does not always equal the moles of product.

Consider

${N}_{2} + 3 {H}_{2} \rightarrow 2 N {H}_{3}$

The equilibrium constant is

$k = \frac{{\left[N {H}_{3}\right]}^{2}}{\left[{N}_{2}\right] {\left[{H}_{2}\right]}^{3}}$ and the units of k would be

([moldm^"-3"]^2)/([moldm^"-3"][moldm^"-3"]^3)=1/(mol^2dm^"-6")